Left Termination proof of /tmp/tmpoGRyb5/intlist.pl

Left Termination of the query pattern int_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

intlist([], []).
intlist(.(X, XS), .(s(X), YS)) :- intlist(XS, YS).
int(0, 0, .(0, [])).
int(0, s(Y), .(0, XS)) :- int(s(0), s(Y), XS).
int(s(X), 0, []).
int(s(X), s(Y), XS) :- ','(int(X, Y, ZS), intlist(ZS, XS)).

Queries:

int(g,g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

The argument filtering Pi contains the following mapping:
int_in(x1, x2, x3) = int_in(x1, x2)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
[] = []
int_out(x1, x2, x3) = int_out(x3)
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3) = U2(x3)
U4(x1, x2, x3, x4) = U4(x4)
intlist_in(x1, x2) = intlist_in(x1)
U1(x1, x2, x3, x4) = U1(x1, x4)
intlist_out(x1, x2) = intlist_out(x2)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

INT_IN(s(X), s(Y), XS) → U3¹(X, Y, XS, int_in(X, Y, ZS))
INT_IN(s(X), s(Y), XS) → INT_IN(X, Y, ZS)
INT_IN(0, s(Y), .(0, XS)) → U2¹(Y, XS, int_in(s(0), s(Y), XS))
INT_IN(0, s(Y), .(0, XS)) → INT_IN(s(0), s(Y), XS)
U3¹(X, Y, XS, int_out(X, Y, ZS)) → U4¹(X, Y, XS, intlist_in(ZS, XS))
U3¹(X, Y, XS, int_out(X, Y, ZS)) → INTLIST_IN(ZS, XS)
INTLIST_IN(.(X, XS), .(s(X), YS)) → U1¹(X, XS, YS, intlist_in(XS, YS))
INTLIST_IN(.(X, XS), .(s(X), YS)) → INTLIST_IN(XS, YS)

The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

The argument filtering Pi contains the following mapping:
int_in(x1, x2, x3) = int_in(x1, x2)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
[] = []
int_out(x1, x2, x3) = int_out(x3)
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3) = U2(x3)
U4(x1, x2, x3, x4) = U4(x4)
intlist_in(x1, x2) = intlist_in(x1)
U1(x1, x2, x3, x4) = U1(x1, x4)
intlist_out(x1, x2) = intlist_out(x2)
INTLIST_IN(x1, x2) = INTLIST_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4) = U4¹(x4)
INT_IN(x1, x2, x3) = INT_IN(x1, x2)
U2¹(x1, x2, x3) = U2¹(x3)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

INT_IN(s(X), s(Y), XS) → U3¹(X, Y, XS, int_in(X, Y, ZS))
INT_IN(s(X), s(Y), XS) → INT_IN(X, Y, ZS)
INT_IN(0, s(Y), .(0, XS)) → U2¹(Y, XS, int_in(s(0), s(Y), XS))
INT_IN(0, s(Y), .(0, XS)) → INT_IN(s(0), s(Y), XS)
U3¹(X, Y, XS, int_out(X, Y, ZS)) → U4¹(X, Y, XS, intlist_in(ZS, XS))
U3¹(X, Y, XS, int_out(X, Y, ZS)) → INTLIST_IN(ZS, XS)
INTLIST_IN(.(X, XS), .(s(X), YS)) → U1¹(X, XS, YS, intlist_in(XS, YS))
INTLIST_IN(.(X, XS), .(s(X), YS)) → INTLIST_IN(XS, YS)

The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

The argument filtering Pi contains the following mapping:
int_in(x1, x2, x3) = int_in(x1, x2)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
[] = []
int_out(x1, x2, x3) = int_out(x3)
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3) = U2(x3)
U4(x1, x2, x3, x4) = U4(x4)
intlist_in(x1, x2) = intlist_in(x1)
U1(x1, x2, x3, x4) = U1(x1, x4)
intlist_out(x1, x2) = intlist_out(x2)
INTLIST_IN(x1, x2) = INTLIST_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4) = U4¹(x4)
INT_IN(x1, x2, x3) = INT_IN(x1, x2)
U2¹(x1, x2, x3) = U2¹(x3)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

INTLIST_IN(.(X, XS), .(s(X), YS)) → INTLIST_IN(XS, YS)

The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

The argument filtering Pi contains the following mapping:
int_in(x1, x2, x3) = int_in(x1, x2)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
[] = []
int_out(x1, x2, x3) = int_out(x3)
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3) = U2(x3)
U4(x1, x2, x3, x4) = U4(x4)
intlist_in(x1, x2) = intlist_in(x1)
U1(x1, x2, x3, x4) = U1(x1, x4)
intlist_out(x1, x2) = intlist_out(x2)
INTLIST_IN(x1, x2) = INTLIST_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

INTLIST_IN(.(X, XS), .(s(X), YS)) → INTLIST_IN(XS, YS)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
.(x1, x2) = .(x1, x2)
INTLIST_IN(x1, x2) = INTLIST_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

INTLIST_IN(.(X, XS)) → INTLIST_IN(XS)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

INTLIST_IN(.(X, XS)) → INTLIST_IN(XS)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

INT_IN(s(X), s(Y), XS) → INT_IN(X, Y, ZS)
INT_IN(0, s(Y), .(0, XS)) → INT_IN(s(0), s(Y), XS)

The TRS R consists of the following rules:

int_in(s(X), s(Y), XS) → U3(X, Y, XS, int_in(X, Y, ZS))
int_in(s(X), 0, []) → int_out(s(X), 0, [])
int_in(0, s(Y), .(0, XS)) → U2(Y, XS, int_in(s(0), s(Y), XS))
int_in(0, 0, .(0, [])) → int_out(0, 0, .(0, []))
U2(Y, XS, int_out(s(0), s(Y), XS)) → int_out(0, s(Y), .(0, XS))
U3(X, Y, XS, int_out(X, Y, ZS)) → U4(X, Y, XS, intlist_in(ZS, XS))
intlist_in(.(X, XS), .(s(X), YS)) → U1(X, XS, YS, intlist_in(XS, YS))
intlist_in([], []) → intlist_out([], [])
U1(X, XS, YS, intlist_out(XS, YS)) → intlist_out(.(X, XS), .(s(X), YS))
U4(X, Y, XS, intlist_out(ZS, XS)) → int_out(s(X), s(Y), XS)

The argument filtering Pi contains the following mapping:
int_in(x1, x2, x3) = int_in(x1, x2)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
[] = []
int_out(x1, x2, x3) = int_out(x3)
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3) = U2(x3)
U4(x1, x2, x3, x4) = U4(x4)
intlist_in(x1, x2) = intlist_in(x1)
U1(x1, x2, x3, x4) = U1(x1, x4)
intlist_out(x1, x2) = intlist_out(x2)
INT_IN(x1, x2, x3) = INT_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

INT_IN(s(X), s(Y), XS) → INT_IN(X, Y, ZS)
INT_IN(0, s(Y), .(0, XS)) → INT_IN(s(0), s(Y), XS)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
0 = 0
.(x1, x2) = .(x1, x2)
INT_IN(x1, x2, x3) = INT_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

INT_IN(0, s(Y)) → INT_IN(s(0), s(Y))
INT_IN(s(X), s(Y)) → INT_IN(X, Y)

From the DPs we obtained the following set of size-change graphs:

INT_IN(s(X), s(Y)) → INT_IN(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
INT_IN(0, s(Y)) → INT_IN(s(0), s(Y))
The graph contains the following edges 2 >= 2